Exercise 1773Linear algebra › Matrices and linear systemsL1Added 3 Nov 2026

Positive definiteness of a symmetric matrix

#positive definite#Sylvester criterion#eigenvalues

Statement

Show that
$A = \begin{pmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{pmatrix}$
is positive definite.

Hint

Use Sylvester's criterion (leading principal minors) or compute the eigenvalues.

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Solution

Leading principal minors.
- $\Delta_1 = 2 > 0$ .
- $\Delta_2 = \begin{vmatrix} 2 & -1 \\ -1 & 2 \end{vmatrix} = 4 - 1 = 3 > 0$ .
- $\Delta_3 = \det A = 2(4-1) - (-1)(-2) = 6 - 2 = 4 > 0$ .

Conclusion. By Sylvester's criterion, $A$ is positive definite.

Alternative (eigenvalues). The eigenvalues are $2$ , $2-\sqrt2$ , $2+\sqrt2$ , all positive. So $A$ is positive definite.

Positive definiteness of a symmetric matrix

Statement

Solution

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